Discrete Calder´ on Reproducing Formula

We will use the classical decomposition on the homogeneous space by M. Christ in [11] and by Sawyer-Wheeden in [61]. Here, we use the statement in [11].

Lemma 2.14 ( [11]). χ is the space of homogeneous type, ∃ {Q^k_τ ⊂ χ : k ∈ Z, τ ∈ Ik} of open subsets, where I_k is some index set, δ ∈ (0, 1), C1, C2> 0, s.t.

(i) µ(χ\ ∪τ Q^k_τ) = 0 for each fixed k and Q^k_α∩ Q^k_β = ∅ if α 6= β.

(ii) Q^l_β ⊂ Q^k_α or Q^l_β∩ Q^k_α= ∅ for l ≥ k.

(iii) ∃ ! β, s.t. Q^k_α⊂ Q^l_β. (iv) diam(Q^k_τ) ≤ C1δ^k;

(v) Q^k_τ contains some ball B(z^k_τ, C₂δ^k).

Also, we denote by Q^k,ντ , ν = 1, 2, . . . , N (k, α), the set of all cubes Q^k+jτ ⊂ Q^k_τ and j is a positive large integer such that

2^−jC1 < 1 3

Denote by z_τ^k,ν the “center” of Q^k,τ_τ and by y_τ^k,ν a point in Q^k,ν_τ .

Now we’re ready to introduce the Discrete Calder´on Reproducing Formula. Recall the dis-crete Riemann sum operator on M,

S(f )(x) =X

k∈N

X

τ ∈Ik

N (k,τ )

X

ν=1

ˆ

Q^k,ντ

D^N_k(y) dyDk(f )(y_τ^k,ν)

where D^N_k =P

|l|≤ND_k+l.

We firstly verify S is well defined and bounded on L²(M ).

Lemma 2.15. There exists some constant C > 0, such that for all y^k,ν_τ ∈ Q^k,ν_τ and f ∈ L²(M ),

X

k∈N

X

τ ∈Ik

N (k,τ )

X

ν=1

µ(Q^k,ν_τ )|D_k(f )(y_τ^k,ν)|²≤ Ckf k²_L2(M )

Proof. By Proposition 2.3, we have

|D_kD_l(z, x)| . 2^−|k−l| (1 + 2^kρ(z, x))⁻¹ Vol(B_(X,d)(z, 2^−k+ ρ(z, x)))

where the implicit constant is independent of k, l, z, x. Notice that for all x ∈ M and any z, y ∈ Q^k,ντ , we have that ρ(y, z) ≤ C₁2^−j2^−k ≤ C₁2^−j(2^−k + ρ(x, y)), where C₁2^−j < 1. Thus,

From this, it follows that

|D_k(f )(y_τ^k,ν)| ≤X

Also, we have

Then, the function defined by

f (x) =X

Proof. By the definition, ˆ

where {(D_k,α^N , 2^−k)|(D_k, 2^−k) ∈ D)} and {(D_k^N0,α, 2^−k0)|(D_k⁰, 2^−k0) ∈ D} are both the bounded sets of elementary operators.

Hence, For each k, τ, ν, k⁰, τ⁰, ν⁰, it follows that

By using Lemma 2.3, the last term can be controlled by

CN0,N,m,D2^{−|k−k0|N0} · µ(Q^k,ν_τ )µ(Q^k_τ⁰0^,ν0) (1 + 2^k∧k0ρ(y^k,ντ , y_τ^k00^,ν0))^−m

× (1 + 2^k∧k0ρ(yτ^k,ν, y_τ^k00^,ν0))^−m

With the Cauchy Schwartz inequality, it’s only left to verify,

X

≤X

k⁰

2^{−|k−k0|(N0−N1)} X

τ⁰∈I_k0

N (k⁰,ν⁰)

X

ν⁰=1

ˆ

Q^k0,ν0

τ 0

(1 + 2^k0ρ(y^k,ντ , y⁰))^−2m

Vol B_(X,d)(y^k,ν_τ , 2^−k0+ ρ(y^k,ν_τ , y⁰))
dµ(y⁰)

≤X

k⁰

2^{−|k−k0|(N0−N1)} ˆ

M

(1 + 2^k0ρ(y_τ^k,ν, y⁰))^−2m

Vol B_(X,d)(yτ^k,ν, 2^−k0 + ρ(yτ^k,ν, y⁰))
dµ(y⁰) ≤ C_m

and similarly, for B, we have

X

k≤k⁰

X

τ ∈I_k N (k,τ )

X

ν=1

µ(Q^k,ντ )(1 + 2^kρ(y_τ^k00^,ν0, yτ^k,ν))^−m

Vol B_(X,d)(y^k_τ⁰0^,ν0, 2^−k+ ρ(y_τ^k0^0,ν0, y^k,ντ ))
≤ X

k≤k⁰

2^{−|k−k0|N0}Cm ≤ C_m

With the above two lemmas, it follows immediately that

Theorem 2.17. Let the notation be the same as above with j satisfying 2^−jC1 < ¹₃. Then the discrete Riemann sum operator S is bounded on L²(M ). That is, there is a constant C > 0, only depending on N , such that for all f ∈ L²(M),

kSf k_L2(M )≤ Ckf k_L2(M )

Proof. From Lemma 2.15 and Lemma 2.16, it follows that

kSf k²_L2(M ) ≤ kX

k∈Z

X

τ ∈Ik

N (k,τ )

X

ν=1

[µ(Q^k,ν_τ )]^−1/2|a^k,τ_τ | ˆ

Q^k,ν_τ

D_k^N(x, y) dyk_L²_{(M )}

≤X

k∈Z

X

τ ∈Ik

N (k,τ )

X

ν=1

|a^k,ν_τ |² ≤ kf k²_L2(M )

where a^k,ν_τ = [µ(Q^k,ν_τ )]^1/2|D_k(f )(y^k,ν_τ )|.

Next, we prove that S is invertible. To do this, we define R = I − S and denote by R(x, y) for its kernel. Actually, we can prove: for any n₀∈ N,

Claim 1: For h ∈ N,

h→∞lim kR^h(f )k_L^p_{(M )}≤ lim

h→∞(C_p,N2^−ε0N+ C_p,m,N,D2^−jε)^hkf k_Lp(M ) = 0

h→∞lim kR^h(f )k_{T (n0,m)}≤ lim

h→∞(C_p,N2^−N0N+ C_p,m,N,D2^−jε)^hkf k_{T (n0,m)}= 0

Claim 2:

kS⁻¹k_Lp(M )→L^p(M )< ∞ kS⁻¹k_{T (n0,m)→T (n0,m)}< ∞

Recall that I = UN+ RN, thus

R(f )(x) = (I − S)(f )(x)

=X

k∈N

X

τ ∈Ik

N (k,τ )

X

ν=1

ˆ

Q^k,ν_τ

D^N_k(x, y)[D_k(f )(y) − D_k(f )(y_τ^k,ν)] dy +X

k∈N

X

|l|>M

D_k+lD_k(f )(x)

≡X

k∈N

G_k(f )(x) + RN(f )(x)

≡ G(f )(x) + R_N(f )(x)

Let G_k(x, y) be the kernel of G_k. We now verify that G_k(x, y) and hence G(x, y) satisfies all the desired estimates. Clearly,

G(x, y) =X

k∈N

X

τ ∈I_k N (k,τ )

X

ν=1

ˆ

Q^k,ντ

D^N_k(x, z)[D_k(z, y) − D_k(y_τ^k,ν, y)] dz

=X

k∈N

G_k(x, y)

For each G_k, we can prove the following,

Lemma 2.17. {(Gk, 2^−k)} is a bounded set of the pre-elementary operators.

Proof. By the construction of dyadic cubes in Lemma 2.14, for any z ∈ Q^k,ν_τ ,

ρ(z, y_τ^k,ν) ≤ C₁2^−(k+j) = C₁2^−j2^−k ≤ C₁2^−j(2^−k+ ρ(y, z))

We recall that j always satisfies 2^−jC1 < ¹₃. Thus, we have

1 2^−k+ ρ(z, yτ^k,ν)

≤ C 1

2^−k+ ρ(y, z) (2.3)

and

1

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(z, yτ^k,ν)))
≤ C 1

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(y, z)))
(2.4)

Also by the definition of elementary operators, ∀m, ∃Cm,D= C(m, D), s.t.

|D_k(z, y) − D_k(y_τ^k,ν, y)| ≤ max

z^∗∈Q^{N (k,ν)}_τ

|γ|=1

 

(W_x)^γD_k(z^∗, y)  

ρ(z, y^k,ν_τ )

≤ C_m,D max

z^∗∈Q^k,ντ

2^k (1 + 2^kρ(z^∗, y))^−m

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(z^∗, y)))
ρ(z, y

k,ν τ )

≤ C_m,DC₁2^−j (1 + 2^kρ(y, z))^−m

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(y, z)))

where the last inequality comes from (2.3), (2.4), ρ(z, yτ^k,ν) ≤ C12^−j−k and |γ| = 1.

Consequently,

|G_k(x, y)| =   

X

τ ∈Ik

N (k,ν)

X

ν=1

ˆ

Q^k,ν_τ

D^N_k(x, z)[D_k(z, y) − D_k(y_τ^k,ν, y)] dµ(z)   

≤ C_m,N,DC₁2^−j X

τ ∈Ik

N (k,ν)

X

ν=1

ˆ

Q^k,ν_τ

(1 + 2^kρ(x, z))^−m

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(x, z)))

× (1 + 2^kρ(z, y))^−m

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(z, y)))
dµ(z )

≤ C_m,N,DC₁2^−j (1 + 2^kρ(x, y))^−m

Vol B_(X,d)(y, 2^−k(1 + 2^kρ(x, y)))

where the last inequality comes from Lemma 2.3 and C_m,N,D only depends on m, N and D.

Hence, {(G_k, 2^−k)|k ∈ N} is a bounded set of pre-elementary operators.

In fact, we can furthermore obtain the result as follows:

Lemma 2.18. {(G_k, 2^−k)|k ∈ N} is a bounded set of the elementary operators.

Proof. We’ve verify that {(G_k, 2^−k)|k ∈ N} is a bounded set of pre-elementary operators. The result will follow once we show for k ∈ N, we have Gk is a sum of derivatives of operators of the same form, as in the definition of elementary operators. But we have, using Proposition 2.1,

Gk = X

therefore,

G_k= X

|α|,|β|≤1

2−(2−|α|−|β|)k(2^−kW )^αn X

τ ∈Ik

N (k,ν)

X

ν=1

ˆ

Q^k,ν_τ

D_k,α^N (x, z)

×[D_k,β(z, y) − Dk,β(y^k,ν_τ , y)] dµ(z) o

(2^−kW )^β

= X

|α|,|β|≤1

2−(2−|α|−|β|)k(2^−kW )^αG_k,α,β(2^−kW )^β

This completes the proof, since G_k,α,β is of the same form as G_k.

Lemma 2.19. G is a Calder´on-Zygmund operator of order 0. Moreover, ∃ε > 0, s.t.

kGk_Lp(M )→L^p(M ) ≤ C_m,D,N2^−jε

and for any n0 ∈ N,

kGk_{T (n0,m)→T (n0,m)}≤ C_m,D,N2^−jε

Proof. According to the characterization in Theorem 2.11, it follows that G = P

k∈NGk is a Calder´on-Zygmund operator of order 0. Hence, G is L^p(M ) → L^p(M ) bounded, i.e. ∀p > 0,

∃C_p, such that

kGk_Lp(M )→L^p(M ) ≤ C_p

Also note that, for all k, l ∈ N, we have

kG_kG^∗_lk_L2(M )→L²(M )≤ C_m,D,N2^−j2^−|k−l|, kG_lG^∗_kk_L2(M )→L²(M ) ≤ C_m,D,N2^−j2^−|k−l|

The Colter-Stein Lemma now shows

kGk_L2(M )→L²(M )≤ C_m,D,N2^−j

Together, by the interpolation, we have

kGk_Lp(M )→L^p(M ) ≤ C_p,m,D,N2^−jε

On the other hand, ∀f ∈ T (n0, m) and x ∈ M ,

|W^αG(f )(x)| ≤X

k∈N

2^k|α||(2^−kW )^αG_k(f )(x)|

By using the Proposition 2.1, we know that {((2^−kW )^αGk(f ), 2^−k) | (Gk, 2^−k), ∈ D} is a bounded set of elementary operators. For the simplicity, we denote the above new set as {(E_k, 2^−k) | (E_k, 2^−k)

∈ E}.

Thus, we can rewrite the inequality as

|W^αG(f )(x)| ≤X

k∈N

2^k|α||E_k(f )(x)|

Again, applying the similar proof as Lemma 2.8, we have ∀N₁, m,

|W^αG(f )(x)| . Cm,D,N2^−jX

k∈N

2^k|α|2^−N1k 1 + ρ(x1, x)
−m

Vol B_(X,d)(x1, 1 + ρ(x1, x))
kf kT (n0,m)

By the convergence of geometric series, it suffices to verify

X

k∈N

2^k(|α|−N1). 1

which is obviously by taking N₁ large. Therefore , we obtain that

|W^αG(f )(x)| ≤ C_m,D,N2^−j 1 + ρ(x1, x)
−m

Vol B_(X,d)(x₁, 1 + ρ(x₁, x))
kf kT (n0,m)

Lemma 2.20. Let S be the discrete Riemann sum operator on M and R = I − S. Then R is L^p(M ) bounded, i.e.

kRk_Lp(M )→L^p(M ) ≤ C_p2^−ε0N + Cp,m,D,N2^−jε

and R is bounded on T (n₀, m) for any n₀ ∈ N, i.e.

kRk_{T (n0,m)→T (n0,m)}≤ C2^−N0N + Cm,D,N2^−j

Proof. By Lemma 2.9 and Lemma 2.19, it follows that

kRk_Lp(M )→L^p(M ) ≤ kGk_Lp(M )→L^p(M )+ kRNk_Lp(M )→L^p(M )

≤ C_p,N2^−0N + C_p,m,D,N2^−jε

Also,

kRk_{T (n0,m)→T (n0,m)}≤ kGk_{T (n0,m)→T (n0,m)}+ kR_Nk_{T (n0,m)→T (n0,m)}

≤ C_N2^−N0N+ C_m,D,N2^−j

Since N and j are arbitrary, we can choose them large enough such that

C_p,N2^−0N+ C_p,m,D,N2^−jε< 1

and

C_p,N2^−N0N + C_m,D,N2^−j < 1

which implies our Claim 1.

Our Claim 2 follows from the corollary below.

Corollary 2.18. Let S be the discrete Riemann operator on M. Let N, j ∈ N such that

Cp,N2^−0N+ Cp,m,D,N2^−jε< 1

and

CN2^−N0N + Cm,D,N2^−j < 1

hold. Then S has a bounded inverse in L^p(M ) for p ∈ (1, ∞). Namely, there exists a constant C > 0 depending on p such that

kS⁻¹k_Lp(M )→L^p(M ) ≤ C

and for any n0∈ N,

kS⁻¹k_{T (n0,m)→T (n0,m)}≤ C

To establish the discrete Calder´on reproducing formula, we still need the following technical lemma.

Lemma 2.21. Let j satisfy C₁2^−j < ¹₃. For k ∈ Z, any fixed y^k,ντ ∈ Q^k,ν_τ with τ ∈ I_k and

ν = {1, . . . , N (k, τ )}, and any x ∈ M , let

≤ C_D,α,β,m,N     

ˆ

M

(1 + 2^kρ(x, z))^−m Vol B_(X,d)(x, 2^−k+ ρ(x, z))

(1 + 2^kρ(z, y))^−m

Vol B_(X,d)(z, 2^−k+ ρ(z, y))
dµ(z )     

≤ C_D,α,β,m,N (1 + 2^kρ(x, y))^−m Vol B_(X,d)(x, 2^−k+ ρ(x, y))

Therefore, {(H_k, 2^−k) | (D_k, 2^−k) ∈ D} is a bounded set of pre-elementary operators.

Next, we will show for k ∈ N, we have Hk is a sum of derivatives of operators of the same form, as in the definition of elementary operators. But we have, using Proposition 2.1,

H_k(x, z) = X

|α|,|β|≤1

X

τ ∈Ik

N (k,τ )

X

ν=1

ˆ

Q^k,ν_τ

2−(2−|α|−|β|)k(2^−kW )^αD_k,α^N (x, z) dµ(z)

×D_k,β(y_τ^k,ν, y)(2^−kW )^β

where {(D^N_k,α, 2^−k), (D_k,β, 2^−k)|(D_k, 2^−k) ∈ D} is a bounded set of elementary operators. And therefore,

H_k(x, y) = X

|α|,|β|≤1

2−(2−|α|−|β|)k(2^−kW )^αn X

τ ∈Ik

N (k,τ )

X

ν=1

ˆ

Q^k,ντ

D_k,α^N (x, z) dµ(z)

×D_k,β(y_τ^k,ν, y)o

(2^−kW )^β

= X

|α|,|β|≤1

2−(2−|α|−|β|)k(2^−kW )^αH_k,α,β(2^−kW )^β

This completes the proof, since Hk,α,β is of the same form as Hk.

Now let’s prove the Discrete Calder´on Reproducing Formula.

Theorem 2.19 (Discrete Calder´on Reproducing Formula). For any fixed j ∈ N, there exists a family of linear operators { eD_k}_k∈N such that for any fixed yτ^k,ν ∈ Q^k,ν_τ with k ∈ N, τ ∈ Ik, and ν = 1, . . . , N (k, τ ), and all f ∈ C^∞(M ),

f (x) =

∞

X

k=0

X

τ ∈Ik

N (k,τ )

X

ν=1

ˆ

Q^k,ντ

De_k(x, y) dµ(y)D_k(f )(y_τ^k,ν)

=

To finish the proof of the theorem, we still need to verify that for all f ∈ T (n₀, m), a bounded set of elementary operators. For the simplicity, we denote the above new set as {(E_k, 2^−k) |

Next, applying the similar proof method as Lemma 2.8, ∀N₁, m,



Hence, it suffices to verify

X

k>L

2^k|α|2^−N1k . 1

By taking N1 large, this is a geometric sum, and therefore bounded by a constant times its largest term. There exists some θ > 0, such that

X

k>L

2^k|α|2^−N1k . 2^−θL.

Combing all of the above, it follows that



Combing Corollary 2.18, it’s easy to obtain (2.5). Next, let’s consider (2.6).

For L ∈ N, let TL be the operator associated with the kernel

K_L(x, y) = X

where eD_l for l ∈ N are as in Theorem 2.13. We have the following orthogonality estimate:

|D_kDe_l(y^k,ν_τ , z)| . 2^−|k−l| (1 + 2^k∧lρ(y, z))^−m

Vol B_(X,d)(y, 2^−k∧l(1 + 2^k∧lρ(y, z)))
,

|(D^N_k)^∗De_l(y, z)| . 2^−|k−l| (1 + 2^k∧lρ(y, z))^−m

Vol B_(X,d)(y, 2^−k∧l(1 + 2^k∧lρ(y, z)))
.

where the implicit constant is independent of k, l, y_τ^k,ν, z. Besides, we have

X

Hence, by using the Fefferman-Stein’s vector-valued maximal function inequality, and the duality argument, we have

. 2^−L/2

In document Hardy Space Theory And Endpoint Estimates For Multi-Parameter Singular Radon Transforms (Page 27-46)